# Two points on a curve with a common Tangent

1. Jul 29, 2009

### Samuelb88

1. The problem statement, all variables and given/known data
Find the two points on the curve $$y=x^4-2x^2-x$$ that have a common tangent line.

2. Relevant equations
See below.

3. The attempt at a solution

$$y=f(x)=x^4-2x^2-x$$

$$\frac{d}{dx}\right((x^4-2x^2-x)=4x^3-4x-1$$

To my understanding, two points $$A,B$$ who have a common Tangent such that $$T_A=T_B$$

-$$A(a,f(a))$$
-$$B(b,f(b))$$

$$T_A: y=(4a^3-4a-1)x-3a^4+2a^2$$
$$T_B: y=(4b^3-4b-1)x-3b^4+2b^2$$

By setting $$T_A=T_B$$:

$$x(4a^3-4b^3-4a+4b)=3(a^4-b^4)-2(a^2-b^2)$$

--> $$x(4(a^3-b^3)-4(a-b))=3(a^2-b^2)(a^2+b^2)-2(a^2-b^2)$$

--> $$x((a-b)(4(a^2+ab+b^2)-4)=(a-b)(3(a+b)(a^2+b^2)-2(a+b))$$

--> $$x=\frac{3(a+b)(a^2+b^2)-2(a+b)}{4(a^2+ab+b^2)-4)}\right(\$$

Here is where I get lost. I've spent hours re-arranging the numerator and denominator through basic arithmetic operations and trying to get the denominator to cancel out in some way but to no avail. I've spent the last couple of days trying to solve this problem in various ways which is quite embarrassing because the problem seems so simple.

I've also trying setting $$T_A=T_B$$ and setting the equation equal to zero:

$$x(4a^3-4b^3-4a+4b)-3(a^4-b^4)+2(a^2-b^2)=0$$

--> $$x(a-b)(4(a^2+ab+b^2)-4)-(a-b)(3(a+b)(a^2+b^2)+2(a+b))=0$$

Here, from factoring the term "(a-b)" and finding the zeroes, i understand a=b, therefore a-b=0, but by substituting the term (a-b) with 0, i get 0. By substituting all a or b terms with the other, i get x=a thus x=b since a=b.

--> $$x(4(a^2+ab+b^2)-4)-(3(a+b)(a^2+b^2)+2(a+b))=0$$

--> $$2x(2(a^2+ab+b^2)-2)-(a+b)(3(a^2+b^2)+2)=0$$

?:(

Last edited: Jul 29, 2009
2. Jul 29, 2009

### snipez90

All right, I finally got this one, I think. Anyways, my solution is probably not the quickest, but I'll provide an outline and try to find a quicker solution later, maybe.

Let a and b be the points we are trying to determine. Assume a and b are distinct. From f'(a) = f'(b), we get an equality relating a and b (and the number 1). Denote this equality as ***

Without loss of generality, assume b > a. Then [f(b)-f(a)]/(b-a) = f'(b) = f'(a). After lots of factoring and simplification, I got [f(b)-f(a)]/(b-a) = f'(b) down to expressions where the only thing I could use was the equality ***. Applying that twice, I got an extremely simple equality. In fact, you should be able to derive *** and even guess what a and b are. But yeah, from that point onwards, there are probably quicker ways to actually find a and b.

3. Jul 29, 2009

### tiny-tim

Hi Samuelb88!

You were doing fine until you got to TA = TB. :tongue2:

But you've completely misunderstood what that means.

For the lines to be the same, the x terms must be the same (that's one equation) and the constant terms must be the same (that's a separate equation).

Not one big equation!!

4. Jul 31, 2009

### Samuelb88

Still having troubles finding the values or a and b. :(

Alright, so if the slopes are one equation, and the constant (b) is a separate equation, here's what i've worked out.

$$y=x^4-2x^2-x$$

From differentiating w.r.t.x., i get:

$$y'=4x^3-4x-1$$

The equation of the slopes of the tangent at points A(a,f(a)), B(b,f(b)) which are equal:

$$4a^3-4a-1=4b^3-4b-1$$
$$= 4(a^3-b^3)-4(a-b)=0$$
$$= 4(a-b)(a^2+ab+b^2)-4(a-b)=0$$

which leaves me with:

$$4(a-b)(a^2+ab+b^2-1)=0$$

Thus, a-b=0; therefore a=b.

And, for $$a^2+ab+b^2-1=0$$, would be true if a=1 and b=-1. But if a=1, b=-1, then a≠b.

The equation of the constant (b) of the tangent at points A, B which are equal:

$$-3a^4+2a^2=-3b^4+2b^2$$
$$= -3(a^4-b^4)+2(a^2-b^2)=0$$
$$= (a-b)(-3(a^3+a^2b+ab^2+b^3)+2(a+b))=0$$

Again, a-b=0; a=b. And for $$-3(a^3+a^2b+ab^2+b^3)+2(a+b)=0$$, would be true again if a=1 and b=-1 then a≠b.

I'm confused as to how to prove a=1, b=-1 or if i'm even understanding the problem correctly. :|

------------------------------------

Alternatively, if a=b, then from the equation of the slopes at points A,B which are equal:

$$4a^3-4a-1=4b^3-4b-1$$
$$=4a(a^2-1)=4b(b^2-1)$$ /a=b
$$=4a(a^2-1)-4b(a^2-1)=0$$ /factor
$$=(a^2-1)(a-b)=0$$

Therefore, a= ±1; and a=b; then b= ±1.

Thus:
$$x_1=1: y'=-1$$

And:
$$x_2=-1: y'=-1$$

Then:
$$-1=4x^3-4x-1$$
$$0=4x(x^2-1)$$

Therefore:
$$x_1=0, x_2=1, x_3=-1$$

Therefore, at the points A(1,-2), B(-1,0), the curve has a common tangent.

Last edited: Jul 31, 2009
5. Jul 31, 2009

### snipez90

Actually, you missed a solution in setting the constants equal. I essentially reached the same equation by a slightly longer route. Note that the assumption a =/= b is an important one to make in this problem. To summarize your steps:

We get

$$(a-b)(a^2+ab+b^2-1)=0$$

by setting the slopes equal to each other. Assuming that a =/= b gives

$$(a^2+ab+b^2-1)=0$$

By setting the constants equal,

$$(a-b)(-3(a^3+a^2b+ab^2+b^3)+2(a+b))=0$$

Since a =/= b, we have

$$2(a+b)-3(a^3+a^2b+ab^2+b^3)=0 \Rightarrow$$
$$2(a+b)-3(a+b)(a^2 + b^2)=0 \Rightarrow$$
$$(a+b)(2-3(a^2 + b^2))=0 \Rightarrow$$
$$(a+b)((2a^2 + 2b^2 + 2ab)-3(a^2 + b^2))=0 \Rightarrow$$
$$-(a+b)(a^2 -2ab + b^2)=0 \Rightarrow$$
$$(a+b)(a-b)^2 =0 \Rightarrow,$$

where we substituted the equality $$2(a^2+ab+b^2) = 2$$ from above. Since a =/= b, it follows that a + b = 0, implying that a = -b, and the rest is straightforward.

6. Jul 31, 2009

### tiny-tim

Hi Samuelb88!

ooh … snipez90 just beat me to it!

First, the tangents at A and B have to be the same if A and B are the same!

So a = b was always going to be a solution (to TA = TB), but it's a solution you have to ignore!

ok, so that leaves two simultaneous equations:

a2 + ab + b2 = 1

3(a4 - b4) = 2(a2 - b2)

At this point, you don't seem to have noticed that you can factor out (a2 - b2) = (a-b)(a+b).

(If you do that, and then eliminate (a2 + b2), you actually get the inflection points … but they didn't ask you for them, so as it happens it doesn't matter. ).

So a = -b, and you quickly get ±1.